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Chapter 5: Problem 2

The force of static friction acts only between surfaces at rest. Yet that force is essential in walking and in accelerating or braking a car. Explain.

Short Answer

Expert verified
Static friction is essential in walking and in accelerating or braking a car as it prevents the slipping of surfaces in contact. While walking, it is this force that allows the person to be propelled forward when the foot pushes the ground backward. Similarly, in a car, static friction enables the vehicle to move forward when the wheels push backward on the road during acceleration, or to slow down when the brakes exert a forward force on the road during braking.

Step by step solution

01

Understanding the role of Static Friction in Walking

In order for a person to walk, they must first exert a backward force on the ground with their foot. This action pushes the ground backward and, thanks to Newton's third law, the ground also exerts an equal and opposite force on the foot, enabling the person to move forward. Despite appearances, the foot and the surface are not moving relative to each other at the point of contact. Thus, static friction comes into play here. Static friction provides the ground reaction force which stops the foot from slipping and allows effective walking.
02

Understanding the role of static friction in Driving

Similar principles are at play in accelerating or braking a car. When a car accelerates, the wheels push backward on the road (much like a walker's foot), and the road reacts by providing a forward static frictional force. This reaction force pushes the car forward. Conversely, when braking, the brakes exert a force on the wheels, causing them to exert a forward force on the road. The road then reacts, providing a backward static frictional force. As long as the tires don't skid, this interaction is governed by static friction. This force slows the car down and eventually brings it to rest, by opposing the motion of the vehicle.

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Most popular questions from this chapter

Two forces act on a 3.1 -kg mass that undergoes acceleration \(\vec{a}=0.91 \hat{\imath}-0.27 \hat{\jmath} \mathrm{m} / \mathrm{s}^{2} .\) If one force is \(-1.2 \hat{\imath}-2.5 \hat{\jmath} \mathrm{N},\) what's the other?

Two blocks rest on slopes of unequal angles, connected by a rope passing over a pulley (Fig. 5.30). If the blocks have equal masses, will they remain at rest? Why? Neglect friction.

You try to move a heavy trunk, pushing down and forward at an angle of \(50^{\circ}\) below the horizontal. Show that, no matter how hard you push, it's impossible to budge the trunk if the coefficient of static friction exceeds 0.84.

An astronaut is training in an earthbound centrifuge that consists of a small chamber whirled horizontally at the end of a \(5.1-\mathrm{m}-\) long shaft. The astronaut places a notebook on the vertical wall of the chamber and it stays in place. If the coefficient of static friction is \(0.62,\) what's the minimum rate at which the centrifuge must be revolving?

A \(14.6-\mathrm{kg}\) monkey hangs from the middle of a massless rope, each half of which makes an \(11.0^{\circ}\) angle with the horizontal. What's the rope tension? Compare with the monkey's weight.
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